Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{n^2 - 14n + 40}{8n^2 - 40n} \div \dfrac{-8n + 80}{7n - 35} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{n^2 - 14n + 40}{8n^2 - 40n} \times \dfrac{7n - 35}{-8n + 80} $ First factor the quadratic. $p = \dfrac{(n - 10)(n - 4)}{8n^2 - 40n} \times \dfrac{7n - 35}{-8n + 80} $ Then factor out any other terms. $p = \dfrac{(n - 10)(n - 4)}{8n(n - 5)} \times \dfrac{7(n - 5)}{-8(n - 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (n - 10)(n - 4) \times 7(n - 5) } { 8n(n - 5) \times -8(n - 10) } $ $p = \dfrac{ 7(n - 10)(n - 4)(n - 5)}{ -64n(n - 5)(n - 10)} $ Notice that $(n - 5)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 7\cancel{(n - 10)}(n - 4)(n - 5)}{ -64n(n - 5)\cancel{(n - 10)}} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $p = \dfrac{ 7\cancel{(n - 10)}(n - 4)\cancel{(n - 5)}}{ -64n\cancel{(n - 5)}\cancel{(n - 10)}} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $p = \dfrac{7(n - 4)}{-64n} $ $p = \dfrac{-7(n - 4)}{64n} ; \space n \neq 10 ; \space n \neq 5 $